Question: The surface area of a sphere is increasing at a rate of $14\pi$ square meters per hour. At a certain instant, the surface area is $36\pi$ square meters. What is the rate of change of the volume of the sphere at that instant (in cubic meters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $(\sqrt{14\pi})^{^3}$ (Choice B) B $21\pi$ (Choice C) C $36\pi$ (Choice D) D $\dfrac{7}{12}$ The surface area of a sphere with radius $r$ is $4\pi r^2$. The volume of a sphere with radius $r$ is $\dfrac{4}{3}\pi r^3$.
Solution: Setting up the math Let... $r(t)$ denote the sphere's radius at time $t$, $V(t)$ denote the sphere's volume at time $t$, and $S(t)$ denote the sphere's surface area at time $t$. We are given that $S'(t)=14\pi$, We are also given that $S(t_0)=36\pi$ for a specific time $t_0$. We want to find $V'(t_0)$. Relating the measures $S(t)$ and $r(t)$ relate to each other through the formula for the surface area of a sphere: $S(t)=4\pi[r(t)]^2$ We can differentiate both sides to find an expression for $S'(t)$ : $S'(t)=8\pi r(t)r'(t)$ $V(t)$ and $r(t)$ relate to each other through the formula for the volume of a sphere: $V(t)=\dfrac{4}{3}\pi[r(t)]^3$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=4\pi[r(t)]^2r'(t)$ Using the information to solve Let's plug ${S(t_0)}={36\pi}$ into the expression for $S(t_0)$ : $\begin{aligned} {S(t_0)}&=4\pi[r(t_0)]^2 \\\\ {36\pi}&=4\pi[r(t_0)]^2 \\\\ 9&=[r(t_0)]^2 \\\\ {3}&={r(t_0)} \end{aligned}$ Let's plug ${S'(t_0)}={14\pi}$ and ${r(t_0)}={3}$ into the expression for $S'(t_0)$ : $\begin{aligned} {S'(t_0)}&=8\pi{r(t_0)} r'(t_0) \\\\ {14\pi}&=8\pi({3})r'(t_0) \\\\ C{\dfrac{7}{12}}&=C{r'(t_0)} \end{aligned}$ Now let's plug ${r(t_0)}={3}$ and $C{r'(t_0)}=C{\dfrac{7}{12}}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=4\pi[{r(t_0)}]^2C{r'(t_0)} \\\\ &=4\pi({3})^2\left(C{\dfrac{7}{12}}\right) \\\\ &=21\pi \end{aligned}$ In conclusion, the rate of change of the volume of the sphere at that instant is $21\pi$ cubic meters per hour. Since the rate of change is positive, we know that the volume is increasing.